Data anaysis

# load data from previously processed data

import numpy as np
import pandas as pd

df_data = pd.read_csv('data1019.csv')
df_data.head()

	CMPLNT_NUM	CMPLNT_DATE	CMPLNT_YEAR	CMPLNT_MONTH	OFNS_DESC	BORO_NM	Latitude	Longitude
0	408450072	2012-01-02	2012	1	OFF. AGNST PUB ORD SENSBLTY &	BROOKLYN	40.597121	-73.941249
1	114547452	2012-01-02	2012	1	BURGLARY	QUEENS	40.762102	-73.820334
2	579666649	2012-01-02	2012	1	ARSON	BRONX	40.851316	-73.902117
3	228510203	2012-01-02	2012	1	PETIT LARCENY	QUEENS	40.709501	-73.786483
4	329522444	2012-01-02	2012	1	BURGLARY	BROOKLYN	40.668598	-73.889725

Question 1. What are the most common crime types? Which borough had the most crime number?

crime_group_df = pd.DataFrame({'Number of Crime': df_data.groupby(['OFNS_DESC']).size()}).reset_index()
df_data['OFNS_DESC'].value_counts()

PETIT LARCENY                      676860
HARRASSMENT 2                      514354
ASSAULT 3 & RELATED OFFENSES       424031
CRIMINAL MISCHIEF & RELATED OF     380670
GRAND LARCENY                      352753
                                    ...  
FORTUNE TELLING                         4
KIDNAPPING AND RELATED OFFENSES         3
OFFENSES AGAINST MARRIAGE UNCL          2
UNDER THE INFLUENCE OF DRUGS            2
LOITERING FOR DRUG PURPOSES             1
Name: OFNS_DESC, Length: 70, dtype: int64

Let's visualize it!

# Bar chart that shows the amount of crime since year 2012
crime_group_df.plot(kind='bar', x='OFNS_DESC', y='Number of Crime', figsize=(20,5))

<AxesSubplot:xlabel='OFNS_DESC'>

# Line fig that shows the amount and trend of each type of crime since the year 2012
import matplotlib.pyplot as plt
fig, ax = plt.subplots(figsize=(20, 20))
df_data.groupby(['CMPLNT_YEAR', 'OFNS_DESC']).count()['CMPLNT_NUM'].unstack().plot(ax=ax)

<AxesSubplot:xlabel='CMPLNT_YEAR'>

Let's find out which borough has the most number of crimes

# most often borough
borough_top_2012 = pd.DataFrame({'count': df_data.groupby(['BORO_NM']).size()}).reset_index()
borough_top_2012.sort_values('count', ascending=False, inplace=True)
borough_top_2012

	BORO_NM	count
1	BROOKLYN	1144279
2	MANHATTAN	926791
0	BRONX	833774
3	QUEENS	764509
4	STATEN ISLAND	174446

---Q1 ANSWER: From the outcome we know that PETIT LARCENY is the most common crime type. And the top 5 crime types are PETIT LARCENY, HARRASSMENT 2, ASSAULT 3 & RELATED OFFENSES, CRIMINAL MISCHIEF & RELATED OF, and GRAND LARCENY.

Question 2. Did some of the efforts of the police system and the social education system work? What is the trend of crimes declined?

Let's visualize the trend

# by year
fig, ax = plt.subplots(figsize=(20, 5))
df_data.groupby(['CMPLNT_YEAR']).count()['CMPLNT_NUM'].plot(ax=ax)

<AxesSubplot:xlabel='CMPLNT_YEAR'>

# by month
fig, ax = plt.subplots(figsize=(20, 5))
df_data.groupby(['CMPLNT_YEAR', 'CMPLNT_MONTH']).count()['CMPLNT_NUM'].plot(ax=ax)

<AxesSubplot:xlabel='CMPLNT_YEAR,CMPLNT_MONTH'>

---Q2 ANSWER: The amount of crime is decreasing year by year! The efforts of the police system and the social education system works fine!

Question 3. What is the crime trend during a year? which month is the peak of crime?

First, let's count the number since 2012

# the total amount by month since 2012
crime_group_mon_df = pd.DataFrame({'Number of Crime': df_data.groupby(['CMPLNT_MONTH']).size()}).reset_index()
crime_group_mon_df.sort_values('Number of Crime', ascending=False, inplace=True)
crime_group_mon_df.reset_index(drop=True)

	CMPLNT_MONTH	Number of Crime
0	8	347684
1	7	347652
2	5	339394
3	10	335357
4	6	332877
5	9	331210
6	4	312878
7	3	311535
8	1	305770
9	12	304898
10	11	302668
11	2	274565

crime_group_mon_df.plot.scatter(x='CMPLNT_MONTH',
                      y='Number of Crime',
                      c='DarkBlue')

<AxesSubplot:xlabel='CMPLNT_MONTH', ylabel='Number of Crime'>

# the total amount by year and month since 2012
crime_group_mon_df2 = pd.DataFrame({'Number of Crime': df_data.groupby(['CMPLNT_YEAR','CMPLNT_MONTH']).size()}).reset_index()
crime_group_mon_df2.reset_index(drop=True)
crime_group_mon_df2.head()

	CMPLNT_YEAR	CMPLNT_MONTH	Number of Crime
0	2012	1	39802
1	2012	2	39218
2	2012	3	43220
3	2012	4	41631
4	2012	5	44882

crime_group_mon_df2.plot.scatter(x='CMPLNT_MONTH',
                      y='Number of Crime',
                      c='DarkBlue')

<AxesSubplot:xlabel='CMPLNT_MONTH', ylabel='Number of Crime'>

We can conclude from the data and figures that July and August have the most number of crime. Criminals are more active in summer days.

Then, let's focus on the most recent year 2019 for further analysis.

# get the data of 2019
df_data_2019=df_data.loc[df_data['CMPLNT_YEAR'] == 2019]
df_data_2019.reset_index(drop=True)
df_data_2019.head()

	CMPLNT_NUM	CMPLNT_DATE	CMPLNT_YEAR	CMPLNT_MONTH	OFNS_DESC	BORO_NM	Latitude	Longitude
3395512	442809231	2019-01-01	2019	1	SEX CRIMES	BROOKLYN	40.674583	-73.930222
3395513	651750698	2019-01-01	2019	1	GRAND LARCENY	MANHATTAN	40.748831	-73.988478
3395514	394629573	2019-01-01	2019	1	CRIMINAL MISCHIEF & RELATED OF	BRONX	40.827820	-73.925931
3395515	967895740	2019-01-01	2019	1	HARRASSMENT 2	MANHATTAN	40.804384	-73.937422
3395516	293332550	2019-01-01	2019	1	HARRASSMENT 2	BRONX	40.901630	-73.896747

# most often offence type
type_top5 = pd.DataFrame({'count': df_data_2019.groupby(['OFNS_DESC']).size()})
type_top5.sort_values('count', ascending=False, inplace=True)
type_top5.head()

	count
OFNS_DESC
PETIT LARCENY	87644
HARRASSMENT 2	70972
ASSAULT 3 & RELATED OFFENSES	52581
CRIMINAL MISCHIEF & RELATED OF	46314
GRAND LARCENY	41041

# most offten borough
borough_top = pd.DataFrame({'count': df_data_2019.groupby(['BORO_NM']).size()}).reset_index()
borough_top.sort_values('count', ascending=False, inplace=True)
borough_top

	BORO_NM	count
1	BROOKLYN	129370
2	MANHATTAN	113649
0	BRONX	99002
3	QUEENS	90146
4	STATEN ISLAND	18510

# the total amount by month of 2019
crime_group_mon_df_2019 = pd.DataFrame({'Number of Crime': df_data_2019.groupby(['CMPLNT_MONTH']).size()}).reset_index()
crime_group_mon_df_2019.sort_values('Number of Crime', ascending=False, inplace=True)
crime_group_mon_df_2019.reset_index(drop=True)

	CMPLNT_MONTH	Number of Crime
0	7	42045
1	8	40650
2	5	40022
3	9	39878
4	6	39708
5	10	39165
6	3	36284
7	4	36247
8	11	35981
9	1	35843
10	12	33383
11	2	31770

crime_group_mon_df_2019.plot.scatter(x='CMPLNT_MONTH',
                      y='Number of Crime',
                      c='DarkBlue')

<AxesSubplot:xlabel='CMPLNT_MONTH', ylabel='Number of Crime'>

The 2019 data shows the same trend as history

---Q3 ANSWER: Through out the year, the crime has a curve trend. The peak is Jul and Aug, which is the busiest time of the police system.

Question 4. Which area has the most concentrated crime that needs special attention? Is this area related to it's economic prosperity?

Use geopy library to get the latitude and longitude values of New York City.

!conda install -c conda-forge geopy --yes 
from geopy.geocoders import Nominatim # convert an address into latitude and longitude values

Collecting package metadata (current_repodata.json): done
Solving environment: done

# All requested packages already installed.

address = 'New York City, NY'

geolocator = Nominatim(user_agent="ny_explorer")
location = geolocator.geocode(address)
latitude = location.latitude
longitude = location.longitude
print('The geograpical coordinate of New York City are {}, {}.'.format(latitude, longitude))

The geograpical coordinate of New York City are 40.7127281, -74.0060152.

import folium library to visualize the data on maps

! pip install folium==0.5.0
import folium # plotting library

borough_map = folium.Map(location=[40.7127281, -74.0060152], zoom_start=10.5, tiles='Stamen Toner')

Due to the large amount of data, let's focus on Aug of 2019 to get the visualization

# get the data of Aug, 2019
df_data_201908=df_data_2019.loc[df_data['CMPLNT_MONTH'] == 8]
df_data_201908.reset_index(drop=True)
df_data_201908.head()

	CMPLNT_NUM	CMPLNT_DATE	CMPLNT_YEAR	CMPLNT_MONTH	OFNS_DESC	BORO_NM	Latitude	Longitude
3657431	645757844	2019-08-01	2019	8	DANGEROUS WEAPONS	BRONX	40.840188	-73.926463
3657432	815573324	2019-08-01	2019	8	HARRASSMENT 2	BROOKLYN	40.664307	-73.869972
3657433	620131398	2019-08-01	2019	8	CRIMINAL MISCHIEF & RELATED OF	MANHATTAN	40.710229	-74.007746
3657434	831375149	2019-08-01	2019	8	OTHER OFFENSES RELATED TO THEF	BROOKLYN	40.676981	-73.937310
3657435	203426514	2019-08-01	2019	8	HARRASSMENT 2	BRONX	40.844884	-73.891583

We cannot visualize all the cases due to the large amount, so we can cluster them for better visualization.

from folium.plugins import MarkerCluster
MarkerCluster()

marker_cluster = MarkerCluster().add_to(borough_map)

for lat, lon in zip(df_data_201908.Latitude, df_data_201908.Longitude):
    folium.Marker([lat, lon]).add_to(marker_cluster)
borough_map

Define Foursquare Credentials and Version

CLIENT_ID = 'YQ0X3FK5BEZZJ5VJXXTN0WMWV1E533RJMYHEW5NTAK5A5NUL' # your Foursquare ID
CLIENT_SECRET = 'IQDXGDSDBZVY1RGX2J3AZO3C2VTW2WLK05GREQZWW12QZ0TX' # your Foursquare Secret
VERSION = '20180605' # Foursquare API version
LIMIT = 100 # A default Foursquare API limit value

print('Your credentails:')
print('CLIENT_ID: ' + CLIENT_ID)
print('CLIENT_SECRET:' + CLIENT_SECRET)

Your credentails:
CLIENT_ID: YQ0X3FK5BEZZJ5VJXXTN0WMWV1E533RJMYHEW5NTAK5A5NUL
CLIENT_SECRET:IQDXGDSDBZVY1RGX2J3AZO3C2VTW2WLK05GREQZWW12QZ0TX

Search the whole city, get the top 100 venues.

url = 'https://api.foursquare.com/v2/venues/search?near=New%20York,%20NY&client_id={}&client_secret={}&v={}&limit={}&categoryId=4d4b7105d754a06378d81259'.format(
    CLIENT_ID, 
    CLIENT_SECRET, 
    VERSION, 
    LIMIT)
url

'https://api.foursquare.com/v2/venues/search?near=New%20York,%20NY&client_id=YQ0X3FK5BEZZJ5VJXXTN0WMWV1E533RJMYHEW5NTAK5A5NUL&client_secret=IQDXGDSDBZVY1RGX2J3AZO3C2VTW2WLK05GREQZWW12QZ0TX&v=20180605&limit=100&categoryId=4d4b7105d754a06378d81259'

import requests # library to handle requests
from pandas.io.json import json_normalize # tranform JSON file into a pandas dataframe
results = requests.get(url).json()
results

Analize the results, get the location information.

df_results = pd.DataFrame(results['response']['venues'])
df_loc=df_results['location']
df_ll_array=np.asarray(df_loc)
df_ll_array

nearby_venues = json_normalize(df_ll_array) # flatten JSON
nearby_venues.head()

<ipython-input-64-3166e0eb3158>:1: FutureWarning: pandas.io.json.json_normalize is deprecated, use pandas.json_normalize instead
  nearby_venues = json_normalize(df_ll_array) # flatten JSON

	address	crossStreet	lat	lng	labeledLatLngs	postalCode	cc	city	state	country	formattedAddress	neighborhood
0	675 6th Ave	at W 21st St	40.741739	-73.993653	[{'label': 'display', 'lat': 40.74173913980557...	10010	US	New York	NY	United States	[675 6th Ave (at W 21st St), New York, NY 1001...	NaN
1	200 5th Ave	at W 23rd St	40.741987	-73.989913	[{'label': 'display', 'lat': 40.74198696405861...	10010	US	New York	NY	United States	[200 5th Ave (at W 23rd St), New York, NY 1001...	NaN
2	NaN	NaN	40.771785	-73.833839	[{'label': 'display', 'lat': 40.77178493034232...	11356	US	College Point	NY	United States	[College Point, NY 11356, United States]	NaN
3	767 5th Ave	btwn E 58th & E 59th St	40.763733	-73.972744	[{'label': 'display', 'lat': 40.7637325, 'lng'...	10153	US	New York	NY	United States	[767 5th Ave (btwn E 58th & E 59th St), New Yo...	NaN
4	200 Broadway	btwn John & Fulton St	40.710520	-74.008965	[{'label': 'display', 'lat': 40.71051988539366...	10038	US	New York	NY	United States	[200 Broadway (btwn John & Fulton St), New Yor...	NaN

print('{} venues were returned by Foursquare.'.format(nearby_venues.shape[0]))

50 venues were returned by Foursquare.

在地图上显示venue

for lat, lng in zip(nearby_venues.lat, nearby_venues.lng):
    folium.Marker([lat, lng]).add_to(borough_map)
borough_map

Make this Notebook Trusted to load map: File -> Trust Notebook

---Q4 ANSWER: The aera which has the most top 100 venues has the most concentrated crime. Therefore, crime has a direct relationship with economic prosperity.

Question 5. Can the crime trend of 2020 be predicted?

Two ways are used here to predict the number of 2020: 1. Use the historic data grouped by month,then draw a curve to fit the data to figure out the trend during a year. This curve is the average prediction of 2020. __2. Use the

From the trend figure we know that the trend is not linear throughout the year. So polynomial regression methodology is used.

from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
from sklearn.pipeline import Pipeline

def polynomial_model(degree=1):
    polynomial_features = PolynomialFeatures(degree=degree,
                                             include_bias=False)
    linear_regression = LinearRegression(normalize=True)  #normalize=True对数据归一化处理
    pipeline = Pipeline([("polynomial_features", polynomial_features),
                         ("linear_regression", linear_regression)])
    return pipeline

Try the model with degree from 1 to 10 to find the most fit. Get the score and MSE for each degree.

X = np.asarray(crime_group_mon_df2['CMPLNT_MONTH'])
Y = np.asarray(crime_group_mon_df2['Number of Crime'])
X = X.reshape(-1, 1)
Y = Y.reshape(-1, 1)
from sklearn.metrics import mean_squared_error

degrees = np.arange(1,11)
results = []
scores = []
for d in degrees:
    model = polynomial_model(degree=d)
    model.fit(X, Y)
    train_score = model.score(X, Y) 
    mse = mean_squared_error(Y, model.predict(X)) 
    results.append({"model": model, "degree": d, "score": train_score, "mse": mse})
    scores.append(train_score)
for r in results:
    print("degree: {}; train score: {}; mean squared error: {}".format(r["degree"], r["score"], r["mse"]))

degree: 1; train score: 0.06558032532203195; mean squared error: 9506880.509946218
degree: 2; train score: 0.45283034984523385; mean squared error: 5566959.497597453
degree: 3; train score: 0.49697642170506406; mean squared error: 5117812.886574367
degree: 4; train score: 0.5705710496119784; mean squared error: 4369053.680571869
degree: 5; train score: 0.5776630468574382; mean squared error: 4296898.981546839
degree: 6; train score: 0.6172927633491955; mean squared error: 3893702.227946779
degree: 7; train score: 0.6196882748307984; mean squared error: 3869330.051254703
degree: 8; train score: 0.642561285357248; mean squared error: 3636617.7230894826
degree: 9; train score: 0.6433352693265352; mean squared error: 3628743.13171258
degree: 10; train score: 0.6777380689903367; mean squared error: 3278725.55987143

Draw the fitted curve with different degrees.

from matplotlib.figure import SubplotParams

plt.figure(figsize=(12, 14), dpi=200, subplotpars=SubplotParams(hspace=0.3))
for i, r in enumerate(results):
    fig = plt.subplot(5, 2, i+1)
    plt.xlim(0, 13)
    plt.title("LinearRegression degree={}".format(r["degree"]))
    plt.scatter(X, Y, s=5, c='b', alpha=0.5)
    plt.plot(X, r["model"].predict(X), 'r-')

Draw a figure to visualize the scores with different degrees.

find_degree=pd.DataFrame(columns={"degree": '',"score": ''})
find_degree['degree'] = degrees
find_degree['score'] = scores
find_degree

	degree	score
0	1	0.065580
1	2	0.452830
2	3	0.496976
3	4	0.570571
4	5	0.577663
5	6	0.617293
6	7	0.619688
7	8	0.642561
8	9	0.643335
9	10	0.677738

find_degree.plot('degree', 'score')

<AxesSubplot:xlabel='degree'>

Since that a high degree may cause overfitting, we select degree=6.

Let's get the prediction of 2020 with degree 6.

X = np.asarray(crime_group_mon_df2['CMPLNT_MONTH'])
Y = np.asarray(crime_group_mon_df2['Number of Crime'])
X = X.reshape(-1, 1)
Y = Y.reshape(-1, 1)
x2020 = [1,2,3,4,5,6,7,8,9,10,11,12]
x2020 = np.asarray(x2020)
x2020 = x2020.reshape(-1, 1)
model = polynomial_model(degree=6)
model.fit(X, Y)
predict1 = model.predict(x2020)

df_predict1 = pd.DataFrame(columns={"Year2020": '', "Predict": ''})
df_predict1['Year2020'] = [1,2,3,4,5,6,7,8,9,10,11,12]
df_predict1['Predict'] = predict1
df_predict1

	Year2020	Predict
0	1	38093.359328
1	2	34994.544745
2	3	37598.049784
3	4	40332.958507
4	5	41775.122438
5	6	42347.341843
6	7	42741.893211
7	8	43065.402929
8	9	42706.067168
9	10	40923.217964
10	11	38159.235498
11	12	38073.806585

df_predict1.plot('Year2020', 'Predict')

<AxesSubplot:xlabel='Year2020'>

Next, we use only the recent 2 years data to build the model.

crime_group_mon_df3=crime_group_mon_df2.loc[crime_group_mon_df2['CMPLNT_YEAR'] > 2017]
crime_group_mon_df3.reset_index(drop=True)

	CMPLNT_YEAR	CMPLNT_MONTH	Number of Crime
0	2018	1	38220
1	2018	2	34893
2	2018	3	37736
3	2018	4	37863
4	2018	5	41566
5	2018	6	40415
6	2018	7	42436
7	2018	8	42125
8	2018	9	40035
9	2018	10	40574
10	2018	11	38248
11	2018	12	43402
12	2019	1	35843
13	2019	2	31770
14	2019	3	36284
15	2019	4	36247
16	2019	5	40022
17	2019	6	39708
18	2019	7	42045
19	2019	8	40650
20	2019	9	39878
21	2019	10	39165
22	2019	11	35981
23	2019	12	33383

# convert the months to a continues array.
crime_group_mon_df3['mon_count'] = np.arange(1,25)
crime_group_mon_df3

<ipython-input-38-434c4f32c428>:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead

See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
  crime_group_mon_df3['mon_count'] = np.arange(1,25)

	CMPLNT_YEAR	CMPLNT_MONTH	Number of Crime	mon_count
72	2018	1	38220	1
73	2018	2	34893	2
74	2018	3	37736	3
75	2018	4	37863	4
76	2018	5	41566	5
77	2018	6	40415	6
78	2018	7	42436	7
79	2018	8	42125	8
80	2018	9	40035	9
81	2018	10	40574	10
82	2018	11	38248	11
83	2018	12	43402	12
84	2019	1	35843	13
85	2019	2	31770	14
86	2019	3	36284	15
87	2019	4	36247	16
88	2019	5	40022	17
89	2019	6	39708	18
90	2019	7	42045	19
91	2019	8	40650	20
92	2019	9	39878	21
93	2019	10	39165	22
94	2019	11	35981	23
95	2019	12	33383	24

Also, try different degrees

X2 = np.asarray(crime_group_mon_df3['mon_count'])
Y2 = np.asarray(crime_group_mon_df3['Number of Crime'])
X2 = X2.reshape(-1, 1)
Y2 = Y2.reshape(-1, 1)

degrees2 = np.arange(1,11)
results2 = []
scores2 = []

for d in degrees2:
    model2 = polynomial_model(degree=d)
    model2.fit(X2, Y2)
    train_score2 = model2.score(X2, Y2)  #训练集上拟合的怎么样
    mse2 = mean_squared_error(Y2, model2.predict(X2))  #均方误差 cost
    results2.append({"model": model2, "degree": d, "score": train_score2, "mse": mse2})
    scores2.append(train_score2)
for r in results2:
    print("degree: {}; train score: {}; mean squared error: {}".format(r["degree"], r["score"], r["mse"]))

degree: 1; train score: 0.02194053919218486; mean squared error: 8307981.050066424
degree: 2; train score: 0.0936269458116682; mean squared error: 7699051.499658394
degree: 3; train score: 0.10250253935737164; mean squared error: 7623659.086476385
degree: 4; train score: 0.3478417727083193; mean squared error: 5539661.350966508
degree: 5; train score: 0.5637590319809547; mean squared error: 3705584.2111802027
degree: 6; train score: 0.6489015061015418; mean squared error: 2982354.0908300825
degree: 7; train score: 0.6646063110437589; mean squared error: 2848951.9541673847
degree: 8; train score: 0.6720684692960992; mean squared error: 2785565.757481719
degree: 9; train score: 0.7145118083580837; mean squared error: 2425037.107886866
degree: 10; train score: 0.7173964210916087; mean squared error: 2400534.1927909763

from matplotlib.figure import SubplotParams

plt.figure(figsize=(12, 14), dpi=200, subplotpars=SubplotParams(hspace=0.3))
for i, r in enumerate(results2):
    fig = plt.subplot(5, 2, i+1)
    plt.xlim(0, 25)
    plt.title("LinearRegression degree={}".format(r["degree"]))
    plt.scatter(X2, Y2, s=5, c='b', alpha=0.5)
    plt.plot(X2, r["model"].predict(X2), 'r-')

find_degree2=pd.DataFrame(columns={"degree": '',"score": ''})
find_degree2['degree'] = degrees2
find_degree2['score'] = scores2
find_degree2

	degree	score
0	1	0.021941
1	2	0.093627
2	3	0.102503
3	4	0.347842
4	5	0.563759
5	6	0.648902
6	7	0.664606
7	8	0.672068
8	9	0.714512
9	10	0.717396

find_degree2.plot('degree', 'score')

<AxesSubplot:xlabel='degree'>

It is more obvious from this figure that degree 6 is propriate.

Get the prediction with degree 6.

x2020_2 = [13,14,15,16,17,18,19,20,21,22,23,24]
x2020_2 = np.asarray(x2020_2)
x2020_2 = x2020_2.reshape(-1, 1)

model3 = polynomial_model(degree=6)
model3.fit(X2, Y2)
predict2 = model3.predict(x2020_2)

df_predict2 = pd.DataFrame(columns={"Year2020": '', "Predict": ''})
df_predict2['Year2020'] = [1,2,3,4,5,6,7,8,9,10,11,12]
df_predict2['Predict'] = predict2
df_predict2

	Year2020	Predict
0	1	36574.709937
1	2	36247.419442
2	3	36500.919691
3	4	37269.568385
4	5	38384.148948
5	6	39588.681988
6	7	40568.809189
7	8	40991.749621
8	9	40557.828469
9	10	39063.578189
10	11	36476.412079
11	12	33020.870276

df_predict2.plot('Year2020', 'Predict')

<AxesSubplot:xlabel='Year2020'>

compare the 2 ways of prediction

plt.title('Compare 2 methods')
plt.plot(df_predict1['Year2020'], df_predict1['Predict'], color='cyan', label='use historic data')
plt.plot(df_predict1['Year2020'], df_predict2['Predict'], 'b', label='use recent 2 years data')#'b'：color='blue'
plt.legend()  
plt.xlabel('Month')
plt.ylabel('Number')
plt.show()

Apparently, the second prediction is lower than the first one. This is due to the crime number decreasing year by year. Thus, it is believed that the second prediction is more precise.

---Q5 ANSWER: We can use polynomial regression to predict the year 2020. However, there is no actual data to verify the accuracy of the prediction. Therefore, this prediction only gives us a general idea.